# Height of football

1. Jun 6, 2009

### brunettegurl

1. The problem statement, all variables and given/known data

A quarterback throws a football with an initial speed of 19.0 m/s at an angle 30.5deg above the horizontal. Unfortunately, the pass is incomplete and the football drops to the ground, 35.0 m away from the quaterback. From what height was the football released? (no diagram provided)

2. Relevant equations

Vy= Vknoty - gt from which we get ttot= $$\frac{2Vosin\vartheta}{g}$$
from the equation y= Voyty2-0.5gt0.52 we get hmax= v02sin2g$$\vartheta$$/2g
3. The attempt at a solution
so to figure out far it goes i used the equation xmax= v0xttot where it wld now equal vknotcos$$\vartheta$$(2*vknot sin$$\vartheta$$/g) + distance it travelled and i wanted to know what i was doing wrong thanx

2. Jun 6, 2009

### Cyosis

Not sure what you're doing here. You set v_y=0 when the ball touches the ground? This is not true for if there had been no ground it would have continued to fall.

You know the ball traveled a distance of 35 meters in the x-direction. So you can calculate how long it took for the ball to reach that position with $x=v_x t$. Knowing the flight time you can solve the initial height by using $y=y_0+v_0 t-1/2 gt^2$.

3. Jun 6, 2009

### brunettegurl

why wld the velocity be v0 and not Vy??

4. Jun 6, 2009

### Cyosis

It's not I just listed the general kinematic expression for a falling object. In your case the vertical starting velocity is v_y.

Last edited: Jun 6, 2009
5. Jun 6, 2009

thanx